Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskA {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.ni();
            int m = in.ni();
            int[][] arr = new int[n][m];
            int[][] arr2 = new int[n][m];
            for (int i = 0; i < n; ++i)
                arr[i] = in.nia(arr[i], 0, m);
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j)
                    arr2[i][j] = arr[i][j];
            }
            ArrayList<String> ans = new ArrayList<>();
            ArrayList<String> ans2 = new ArrayList<>();
            for (int i = 0; i < n; ++i) {
                int min = Integer.MAX_VALUE;
                for (int j = 0; j < m; ++j)
                    min = Math.min(min, arr[i][j]);
                if (min == 0)
                    continue;
                for (int j = 0; j < m; ++j)
                    arr[i][j] -= min;
                while (min-- > 0)
                    ans.add("row " + (i + 1));
            }
            for (int i = 0; i < m; ++i) {
                int min = Integer.MAX_VALUE;
                for (int j = 0; j < n; ++j)
                    min = Math.min(min, arr[j][i]);
                if (min == 0)
                    continue;
                for (int j = 0; j < n; ++j)
                    arr[j][i] -= min;
                while (min-- > 0)
                    ans.add("col " + (i + 1));
            }
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    if (arr[i][j] != 0) {
                        out.println(-1);
                        return;
                    }
                }
            }
            for (int i = 0; i < m; ++i) {
                int min = Integer.MAX_VALUE;
                for (int j = 0; j < n; ++j)
                    min = Math.min(min, arr2[j][i]);
                if (min == 0)
                    continue;
                for (int j = 0; j < n; ++j)
                    arr2[j][i] -= min;
                while (min-- > 0)
                    ans2.add("col " + (i + 1));
            }
            for (int i = 0; i < n; ++i) {
                int min = Integer.MAX_VALUE;
                for (int j = 0; j < m; ++j)
                    min = Math.min(min, arr2[i][j]);
                if (min == 0)
                    continue;
                for (int j = 0; j < m; ++j)
                    arr2[i][j] -= min;
                while (min-- > 0)
                    ans2.add("row " + (i + 1));
            }
            if (ans.size() < ans2.size()) {
                out.println(ans.size());
                for (String s : ans)
                    out.println(s);
            } else {
                out.println(ans2.size());
                for (String s : ans2)
                    out.println(s);
            }

        }

    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        public String n() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                }
            }
            return st.nextToken();
        }

        public int[] nia(int[] arr, int start, int end) {
            for (int i = start; i < end; ++i)
                arr[i] = ni();
            return arr;
        }

        public int ni() {
            return Integer.parseInt(n());
        }

    }
}